2-Yoneda lemma

Summarize Timeline Top Qs Fact Check

First we define the functor in the opposite direction

${\displaystyle \mu :F(x)\to {\underline {\operatorname {Hom} }}(h_{x},F)}$

as follows. Given an object ${\displaystyle {\overline {x}}}$ in ${\displaystyle F(x)}$, define the natural transformation

${\displaystyle \mu ({\overline {x}}):h_{x}\to F,}$

that is, ${\displaystyle \mu ({\overline {x}})_{y}:\operatorname {Hom} (y,x)\to F(y),}$ by

${\displaystyle \mu ({\overline {x}})_{y}(f)=(Ff){\overline {x}}.}$

(In the below, we shall often drop a subscript for a natural transformation.) Next, given a morphism ${\displaystyle \varphi$ :{\overline {x}}\to x'} in ${\displaystyle F(x)}$, for ${\displaystyle f:y\to x}$, we let ${\displaystyle \mu (\varphi )(f)}$ be

${\displaystyle (Ff)\varphi$ :(Ff){\overline {x}}\to (Ff)x'.}

Then ${\displaystyle \mu (\varphi ):\mu ({\overline {x}})\to \mu (x')}$ is a morphism (a 2-morphism to be precise or a modification in the terminology of Bénabou). The rest of the proof is then to show

1. The above ${\displaystyle \mu }$ is a functor,
2. ${\displaystyle e\circ \mu \simeq \operatorname {id} }$, where ${\displaystyle e}$ is the evaluation at the identity; i.e., ${\displaystyle e(\lambda )=\lambda (\operatorname {id} _{x}),}$ ${\displaystyle e(\alpha$ :\lambda \to \rho )=\alpha (\operatorname {id} _{x}):\lambda (\operatorname {id} _{x})\to \rho (\operatorname {id} _{x}),}
3. ${\displaystyle \mu \circ e\simeq \operatorname {id} .}$

Claim 1 is clear. As for Claim 2,

${\displaystyle e(\mu ({\overline {x}}))=\mu ({\overline {x}})(\operatorname {id} _{x})=F(\operatorname {id} _{x}){\overline {x}}\simeq \operatorname {id} _{F(x)}{\overline {x}}={\overline {x}}}$

where the isomorphism here comes from the fact that ${\displaystyle F}$ is a pseudofunctor. Similarly,${\displaystyle e(\mu (\varphi ))\simeq \varphi .}$ For Claim 3, we have:

${\displaystyle \mu (e(\lambda ))(f)=(Ff\circ \lambda )(\operatorname {id} _{x})\simeq (\lambda \circ h_{x}f)(\operatorname {id} _{x})=\lambda (f).}$

Similarly for a morphism ${\displaystyle \alpha$ :\lambda \to \rho .} ${\displaystyle \square }$

Summarize Timeline Fact Check

Given an ∞-category C, let ${\displaystyle {\widehat {C}}={\underline {\operatorname {Hom} }}(C^{op},{\textbf {Kan}})}$ be the ∞-category of presheaves on it with values in Kan = the ∞-category of Kan complexes. Then the ∞-version of the Yoneda embedding ${\displaystyle C\hookrightarrow {\widehat {C}}}$ involves some (harmless) choice in the following way.

First, we have the hom-functor

${\displaystyle \operatorname {Hom} :C^{op}\times C\to {\textbf {Kan}}}$

that is characterized by a certain universal property (e.g., universal left fibration) and is unique up to a unique isomorphism in the homotopy category ${\displaystyle \operatorname {ho} {\underline {\operatorname {Hom} }}(C\times C^{op},{\textbf {Kan}}).}$^[3][4] Fix one such functor. Then we get the Yoneda embedding functor in the usual way:

${\displaystyle y:C\to {\widehat {C}},\,a\mapsto \operatorname {Hom} (-,a),}$

which turns out to be fully faithful (i.e., an equivalence on the Hom level).^[5] Moreover and more strongly, for each object ${\displaystyle F}$ in ${\displaystyle {\widehat {C}}}$ and object ${\displaystyle a}$ in ${\displaystyle C}$, the evaluation ${\displaystyle e}$ at the identity (see below)

${\displaystyle \operatorname {Hom} (y(a),F)\to F(a)}$

is invertible in the ∞-category of large Kan complexes (i.e., Kan complexes living in a universe larger than the given one).^[6] Here, the evaluation map ${\displaystyle e}$ refers to the composition

${\displaystyle \operatorname {Hom} (y(a),F){\overset {-(a)}{\to }}\operatorname {Hom} (y(a)(a),F(a))=\operatorname {Hom} (\operatorname {Hom} (a,a),F(a))\to F(a)}$

where the last map is the restriction to the identity ${\displaystyle \operatorname {id} _{a}}$.^[7]

The ∞-Yoneda lemma is closely related to the matter of straightening and unstraightening.